Question

A tube has a length of 1.7 cm and a cross-sectional area of 8.0 x 10-4 m2. The tube is filled with a solution of sugar in water. The diffusion constant of sugar in water is 5.0 x 10-10 m2/s. A difference in concentration of 3.5 x 10-3 kg/m3 is maintained between the ends of the tube. How much time is required for 9.0 x 10-13 kg of sugar to be transported though the tube in seconds?

Answer 1

Use Fick's law, the diffusion rate of the water in the tube is as follows At Solve the equation for At. Am C2- DA 2 (Am)L DA(C-C) Here, Am length, A is the area, and D is the diffusion constant Substitute the all known values in the above equation, we get required time is the mass ofthe sucrose C2-G is the difference in concentration, L is the (am)L DA(C2-C) (9.0x10 kg) (0.017 m) 5.0x10-10 ㎡/s)(8.0x10" ㎡)(3.5 x 10-3 kg/m 3 = 10.9 s If rounded off the result to two significant digits Δι=11s